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3 years ago

Which form of radiation is used to directly INCREASE the temperature of water in a nuclear reactor?

Physics

2 answers:

KengaRu [80]3 years ago

7 0

The correct answer to the question is : Gamma radiation.

EXPLANATION:

In a nuclear reactor, the water is heated due to the radiations coming from nuclear fission.

In the radioactive decay of uranium or other radio active substances, three types of radiations are produced which are known as alpha particle, beta particle and gamma ray.

Alpha particle is the double charged helium nucleus while beta is the stream of electrons. Hence, alpha and beta are the material particles. They are not radiations.

The gamma ray is an electromagnetic radiation. Hence, this will be responsible for heating water.

Hence, the correct answer is gamma radiations.

LuckyWell [14K]3 years ago

6 0

The answer would be D because is the strongest form of radiation

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What are facts about balloons?

Nataly_w [17]

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3 years ago

Read 2 more answers

One of your summer lunar space camp activities is to launch a 1130 kg1130 kg rocket from the surface of the Moon. You are a seri

maxonik [38]

Answer:

∆U = 2.296×10^10Joules

Explanation:

Gravitational potential energy is defined as the energy possessed by an object under the influence of gravity due to its virtue of position.

Potential energy U = Fr where;

F is the force of attraction between the masses of the moon and the rocket.

r is the radius or height of the object.

From Newton's law of universal gravitation, F = GMm/r²

Potential energy U = (-GMm/r²)×r

Potential energy U = -GMm/r

The force is negative because the objects act upward.

M is the mass of the rocket

m is the mass of the moon

Gravitational potential energy possessed by the rocket

U1 = -GMm/r1

r1 is the altitude covered by the rocket

Gravitational potential energy possessed by the Moon

U2 = -GMm/(r2+r1)

r2 is the radius of the moon

Change in gravitational potential energy ∆U = U2-U1

∆U = -GMm/(r2+r1)-(-GMm/r1)

∆U = -GMm/(r2+r1) + GMm/r1

∆U = -GMm{1/(r2+r1)-1/r1}

Given

G = 6.67×10^-11m³/kgs²

M = 1130kg

m = 7.36×10²²kg

r1 = 215km = 215,000m

r2 = 1740km = 1,740,000m

∆U = -6.67×10^-11× 7.36×10²² × 1130{1/(215,000+1,740,000)-1/215000}

∆U= -55.47×10¹⁴{1/1955000-1/215000}

∆U = -55.47×10¹⁴{5.12×10^-7 - 4.65×10^-6}

∆U = -284×10^7 + 257.94×10^8

∆U = 22,954,000,000Joules

∆U = 2.296×10^10Joules

8 0

3 years ago

HELP PLS

melisa1 [442]

Answer:

red I think

Explanation:

it's on red so I googled some of it and the closest was red

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3 years ago

Is any force exerted on the vertical sides of the loop that you used in the experiment and how does it affect the apparent mass?

lorasvet [3.4K]

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2 years ago

According to Bernoulli's equation, the pressure in a fluid will tend to decrease if its velocity increases. Assuming that a wind

Pie

Answer:

The pressure drop predicted by Bernoulli's equation for a wind speed of 5 m/s

= 16.125 Pa

Explanation:

The Bernoulli's equation is essentially a law of conservation of energy.

It describes the change in pressure in relation to the changes in kinetic (velocity changes) and potential (elevation changes) energies.

For this question, we assume that the elevation changes are negligible; so, the Bernoulli's equation is reduced to a pressure change term and a change in kinetic energy term.

We also assume that the initial velocity of wind is 0 m/s.

This calculation is presented in the attached images to this solution.

Using the initial conditions of 0.645 Pa pressure drop and a wind speed of 1 m/s, we first calculate the density of our fluid; air.

The density is obtained to be 1.29 kg/m³.

Then, the second part of the question requires us to calculate the pressure drop for a wind speed of 5 m/s.

We then use the same formula, plugging in all the parameters, to calculate the pressure drop to be 16.125 Pa.

Hope this Helps!!!

7 0

3 years ago